Let ln-2n-2n-2n and Ln-2n-2n-3n then as n -A- Both the sequences have limits-B- lim n ln exists but lim n Ln does not exist-C- lim n - ln does not exist but lim n Ln exists-D- Both the sequences do not have limits-

The correct option is C lim_n→∞l_n does not exist but lim _n→∞L_n exists.l_n=2^n+( 2)^n2^n= 1+( 1)^n ⇒ nbsp;lim_n→∞l_n= 0 nbsp; amp; if n is odd 2 ...

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Byju's Answer

Standard XII

Mathematics

Greatest Integer Function

Let ln=2n+-2n...

Question

Let $l_{n} = \frac{2^{n} + (- 2)^{n}}{2^{n}}$ and $L_{n} = \frac{2^{n} + (- 2)^{n}}{3^{n}}$ then as $n \to \infty$

Both the sequences have limits.

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lim n \to \infty l_{n}

exists but

lim n \to \infty L_{n}

does not exist.

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lim n \to \infty l_{n}

does not exist but

lim n \to \infty L_{n}

exists.

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Both the sequences do not have limits.

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Solution

The correct option is C $lim n \to \infty l_{n}$ does not exist but $lim n \to \infty L_{n}$ exists.
$l_{n} = \frac{2^{n} + (- 2)^{n}}{2^{n}} = 1 + (- 1)^{n} \Rightarrow lim n \to \infty l_{n} = {\begin{matrix} 0 & if n is odd 2 & if n is even \end{matrix}$

So, $lim n \to \infty l_{n}$ does not exist.

$L_{n} = {(\frac{2}{3})}^{n} (1 + (- 1)^{n})$
$\Rightarrow lim n \to \infty L_{n} = 0$

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Similar questions

Q. Evaluate:

lim n \to \infty \frac{3^{n} - 2^{n}}{3^{n} + 2^{n}}

Q. The value of

lim n \to \infty \frac{3^{n} + 2^{n}}{3^{n} - 2^{n}}

{lim}_{n \to \infty} (\frac{2^{n + 1} + 3^{n + 1}}{2^{n} + 3^{n}})

is equal to

Q. If

lim n \to \infty (\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + \frac{1}{3 + \sqrt{3 n}} + \dots + \frac{1}{2 n}) = {log}_{e} k

, then

k =

lim n \to \infty \frac{3^{n + 1} + 2^{n + 2}}{3^{n - 1} + 2^{n - 2}}

is equal to

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Let ln=2n+(−2)n2n and Ln=2n+(−2)n3n then as n→∞

The correct option is C limn→∞ln does not exist but limn→∞Ln exists.ln=2n+(−2)n2n=1+(−1)n⇒limn→∞ln={0 if n is odd2 if n is even So, limn→∞ln does not exist. Ln=(23)n(1+(−1)n) ⇒limn→∞Ln=0

Let $l_{n} = \frac{2^{n} + (- 2)^{n}}{2^{n}}$ and $L_{n} = \frac{2^{n} + (- 2)^{n}}{3^{n}}$ then as $n \to \infty$